The impulse calculator applies the impulse-momentum theorem — one of the most practically useful results in classical mechanics — to let you solve for any unknown in the relationship between force, time, mass, and velocity change. Whether you are designing a collision absorber, analysing a sports impact, or working through a kinematics exam question, this tool shows the full step-by-step working so you understand every number.
The physics behind the tool
Impulse (symbol J) is defined as the integral of a net force over the time interval during which it acts. For a constant force this simplifies to:
J = F · Δt
where F is the net force in newtons and Δt is the time interval in seconds. The result has units of N·s (newton-seconds), which are identical to kg·m/s.
Newton’s second law in its original form says that force equals the rate of change of momentum: F = Δp / Δt. Rearranged, this gives the impulse-momentum theorem:
J = Δp = m · v_f − m · v_i
Impulse equals the change in momentum. This is why a 70 kg person decelerating from 10 m/s to rest always receives the same 700 N·s impulse — the only question is how large the force is and over what time. Airbags and crumple zones exploit this by maximising Δt, which minimises the peak force.
Solve-for-variable modes
The calculator supports seven modes, hiding inputs that are not needed:
- Impulse from force — J = F · Δt. Give the applied force and contact time.
- Impulse from momentum change — J = m·v_f − m·v_i. Give mass and both velocities.
- Force — F = J / Δt. Useful for estimating peak collision forces.
- Time interval Δt — Δt = J / F. How long must a pad stay in contact?
- Final velocity v_f — v_f = (J + m·v_i) / m. What speed does a kicked ball reach?
- Initial velocity v_i — v_i = (m·v_f − J) / m. Back-calculate the incoming speed.
- Mass — m = J / (v_f − v_i). Infer mass from measured velocity change and impulse.
Worked example
A 0.145 kg baseball leaves a bat at 40 m/s (in the opposite direction to its arrival at −30 m/s). The bat is in contact for 1 ms (0.001 s).
Using the momentum-change method:
J = m · v_f − m · v_i
J = 0.145 × 40 − 0.145 × (−30)
J = 5.8 − (−4.35)
J = 10.15 N·s
The average force exerted by the bat:
F = J / Δt = 10.15 / 0.001 = 10,150 N (about 1 tonne-force)
That enormous force lasts only a millisecond — a vivid demonstration of why short contact times produce extreme forces for a given impulse.
Formula reference
| Rearrangement | Formula | Solve when you know |
|---|---|---|
| Impulse (force path) | J = F · Δt | Force and contact time |
| Impulse (momentum path) | J = m·v_f − m·v_i | Mass and both speeds |
| Force | F = J / Δt | Impulse and contact time |
| Contact time | Δt = J / F | Impulse and force |
| Final velocity | v_f = (J + m·v_i) / m | Impulse, mass, initial speed |
| Initial velocity | v_i = (m·v_f − J) / m | Impulse, mass, final speed |
| Mass | m = J / (v_f − v_i) | Impulse and both speeds |
All calculations are in SI units. The momentum snapshot panel (visible in momentum modes) shows p_i, p_f, and Δp alongside the main result so you can cross-check both sides of the theorem at once.